## Momentum operator for the Dirac field?

In the borrowed notes I have for last Monday’s lecture (which I missed) I see the momentum operator defined by
\label{eqn:momentumDirac:20}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

There’s a “use Noether’s theorem” comment associated with this. For the scalar field, using Noether’s theorem, we identified the conserved charge of a spacetime translation as the momentum operator
\label{eqn:momentumDirac:40}
P^i = \int d^3 x T^{0i} = – \int d^3 x \pi(x) \spacegrad \phi(x),

and if we plugged in the creation and anhillation operator representation of $$\pi, \phi$$, out comes
\label{eqn:momentumDirac:60}
\BP =
\inv{2} \int \frac{d^3 q}{(2\pi)^3} \Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger},

(plus $$e^{\pm 2 i \omega_\Bp t}$$ terms that we can argue away.)

It wasn’t clear to me how this worked with the Dirac field, but it turns out that this does follow systematically as expected. For a spacetime translation
\label{eqn:momentumDirac:80}
x^\mu \rightarrow x^\mu + a^\mu,

we find
\label{eqn:momentumDirac:100}
\delta \Psi = -a^\mu \partial_\mu \Psi,

so for the Dirac Lagrangian, we have
\label{eqn:momentumDirac:120}
\begin{aligned}
\delta \LL
&= \delta \lr{ \overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \Psi } \\
&=
(\delta \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \delta \Psi \\
&=
(-a^\sigma \partial_\sigma \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } (-a^\sigma \partial_\sigma \Psi ) \\
&=
-a^\sigma \partial_\sigma \LL \\
&=
\partial_\sigma (-a^\sigma \LL),
\end{aligned}

i.e. $$J^\mu = -a^\mu \LL$$.
To plugging this into the Noether current calculating machine, we have
\label{eqn:momentumDirac:160}
\begin{aligned}
\PD{(\partial_\mu \Psi)}{\LL}
&=
\PD{(\partial_\mu \Psi)}{} \lr{ \overline{\Psi} i \gamma^\sigma \partial_\sigma \Psi – m \overline{\Psi} \Psi } \\
&=
\overline{\Psi} i \gamma^\mu,
\end{aligned}

and
\label{eqn:momentumDirac:180}
\PD{(\partial_\mu \overline{\Psi})}{\LL} = 0,

so
\label{eqn:momentumDirac:140}
\begin{aligned}
j^\mu
&=
(\delta \overline{\Psi}) \PD{(\partial_\mu \overline{\Psi})}{\LL}
+
\PD{(\partial_\mu \Psi)}{\LL} (\delta \Psi)
– a^\mu \LL \\
&=
\overline{\Psi} i \gamma^\mu (-a^\sigma \partial_\sigma \Psi)
– a^\sigma {\delta^{\mu}}_{\sigma} \LL \\
&=
– a^\sigma
\lr{
\overline{\Psi} i \gamma^\mu \partial_\sigma \Psi
+ {\delta^{\mu}}_{\sigma} \LL
} \\
&=
-a_\nu
\lr{
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL
}.
\end{aligned}

We can now define an energy-momentum tensor
\label{eqn:momentumDirac:200}
T^{\mu\nu}
=
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL.

A couple things are of notable in this tensor. One is that it is not symmetric, and there’s doesn’t appear to be any hope
of making it so. For example, the space+time components are way different
\label{eqn:momentumDirac:220}
\begin{aligned}
T^{0k} &= \overline{\Psi} i \gamma^0 \partial^k \Psi \\
T^{k0} &= \overline{\Psi} i \gamma^k \partial^0 \Psi,
\end{aligned}

so if we want a momentum like creature, we have to use $$T^{0k}$$, not $$T^{k0}$$. The charge associated with that current is
\label{eqn:momentumDirac:240}
\begin{aligned}
Q^k
&=
\int d^3 x
\overline{\Psi} i \gamma^0 \partial^k \Psi \\
&=
\int d^3 x
\Psi^\dagger (-i \partial_k) \Psi,
\end{aligned}

or translating from component to vector form
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which is the how the momentum operator is first stated in [2]. Here the vector notation doesn’t have any specific representation, but it is interesting to observe how this is directly related to the massless Dirac Lagrangian

\label{eqn:momentumDirac:280}
\begin{aligned}
\LL(m = 0)
&=
\overline{\Psi} i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i (\partial_0 + \gamma_0 \gamma^k \partial_k) \Psi \\
&=
\Psi^\dagger i (\partial_0 – \gamma_0 \gamma_k \partial_k ) \Psi,
\end{aligned}

but since $$\gamma_0 \gamma_k$$ is a $$4 \times 4$$ representation of the Pauli matrix $$\sigma_k$$ Lagrangian itself breaks down into
\label{eqn:momentumDirac:300}
\LL(m = 0)
=
\Psi^\dagger i \partial_0 \Psi
+
\Bsigma \cdot \lr{ \Psi^\dagger (-i\spacegrad) \Psi },

components, and lo and behold, out pops the momentum operator density! There is ambiguity as to what order of products $$\gamma_0 \gamma_k$$, or $$\gamma_k \gamma_0$$ to pick to represent the Pauli basis ([1] uses $$\gamma_k \gamma_0$$), but we also have sign ambiguity in assembling a Noether charge from the conserved current, so I don’t think that matters. Some part of this should be expected this since the Dirac equation in momentum space is just $$\gamma \cdot p – m = 0$$, so there is an intimate connection with the operator portion and momentum.

The last detail to fill in is going from \ref{eqn:momentumDirac:260} to \ref{eqn:momentumDirac:20} using the $$a, b$$ representation of the field. That’s an algebraically messy looking job that I don’t feel like trying at the moment.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## PHY2403H Quantum Field Theory. Lecture 10: Lorentz boosts, generator of spacetime translation, Lorentz invariant field representation. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Lorentz transform symmetries.

From last time, recall that an infinitesimal Lorentz transform has the form
\label{eqn:qftLecture10:20}
x^\mu \rightarrow x^\mu + \omega^{\mu\nu} x_\nu,

where
\label{eqn:qftLecture10:40}
\omega^{\mu\nu} = -\omega^{\nu\mu}

We showed last time that $$\omega^{ij}$$ induces a rotation, and will show today that $$\omega^{0i}$$ is a boost.

We introduced a three index current, factoring out explicit dependence on the incremental Lorentz transform tensor $$\omega^{\mu\nu}$$ as follows
\label{eqn:qftLecture10:80}
J^{\nu \mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^\mu T^{\nu\rho} },

and can easily show that this current has the desired zero four-divergence property
\label{eqn:qftLecture10:100}
\begin{aligned}
\partial_\nu J^{\nu \mu\rho}
&= \inv{2} \lr{
(\partial_\nu x^\rho) T^{\nu\mu}
+
x^\rho {\partial_\nu T^{\nu\mu} }
– (\partial_\nu x^\mu) T^{\nu\rho}
– x^\mu {\partial_\nu T^{\nu\rho} }
} \\
&= \inv{2} \lr{
(\partial_\nu x^\rho) T^{\nu\mu}
– (\partial_\nu x^\mu) T^{\nu\rho}
} \\
&= \inv{2} \lr{
T^{\rho\mu}
+
– T^{\mu\rho}
} \\
&= 0,
\end{aligned}

since the energy-momentum tensor is symmetric.

Defining charge in the usual fashion $$Q = \int d^3 x j^0$$, so we can define a charge for each pair of indexes $$\mu\nu$$, and in particular
\label{eqn:qftLecture10:120}
Q^{0k} = \int d^3 x J^{0 0 k} = \inv{2} \int d^3 x \lr{ x^k T^{00} – x^0 T^{0k} }

\label{eqn:qftLecture10:540}
\begin{aligned}
\dot{Q}^{0k}
&= \int d^3 x \dot{J}^{0 0k} \\
&= \inv{2} \int d^3 x \lr{ x^k \dot{T}^{00} – x^0 \dot{T}^{0k} }
\end{aligned}

However, since $$0 = \partial_\mu T^{\mu \nu} = \dot{T}^{0 \nu} + \partial_j T^{j \nu}$$, or $$\dot{T}^{0 \nu} = -\partial_j T^{j \nu}$$,
\label{eqn:qftLecture10:560}
\begin{aligned}
\dot{Q}^{0k}
&= \inv{2} \int d^3 x \lr{ x^k (-\partial_j T^{j0}) – T^{0k} – x^0 (-\partial_j T^{jk}) } \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0}) + (\partial_j x^k) T^{j0}
– T^{0k} + x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0}) + {T^{k0}}
– {T^{0k}} + x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0})
+ x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x
\partial_j \lr{
-x^k T^{j0}
+ x^0 T^{jk}
},
\end{aligned}

which leaves just surface terms, so $$\dot{Q}^{0k} = 0$$.

### Quantizing:

From our previous identification
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

we have
\label{eqn:qftLecture10:580}
T^{\nu\mu} = \partial^\nu \phi \partial^\mu \phi – g^{\nu\mu} \LL,

or
\label{eqn:qftLecture10:600}
\begin{aligned}
T^{00}
&= \partial^0 \phi \partial^0 \phi – \inv{2} \lr{ \partial_0 \phi \partial^0 \phi + \partial_k \phi \partial^k \phi } \\
&= \inv{2} \partial^0 \phi \partial^0 \phi – \inv{2} (\spacegrad \phi)^2,
\end{aligned}

and
\label{eqn:qftLecture10:620}
T^{0k} = \partial^0 \phi \partial^k \phi,

so we may quantize these energy momentum tensor components as
\label{eqn:qftLecture10:640}
\begin{aligned}
\hatT^{00} &= \inv{2} \hat{\pi}^2 + \inv{2} (\spacegrad \phihat)^2 \\
\hatT^{0k} &= \inv{2} \hat{\pi} \partial^k \phihat.
\end{aligned}

We can now start computing the commutators associated with the charge operator. The first of those commutators is
\label{eqn:qftLecture10:140}
\antisymmetric{\hatT^{00}(\Bx)}{\phihat(\By)}
=
\inv{2}
\antisymmetric{\hat{\pi}^2(\Bx)}{\phihat(\By)},

which can be evaluated using the field commutator analogue of $$\antisymmetric{F(p)}{q} = i F’$$ which is
\label{eqn:qftLecture10:660}
\antisymmetric{F(\hat{\pi}(\Bx))}{\phihat(\By)} = -i \frac{dF}{d \hat{\pi}} \delta(\Bx – \By),

to give
\label{eqn:qftLecture10:680}
\antisymmetric{\hatT^{00}(\Bx)}{\phihat(\By)}
= -i \delta^3(\Bx – \By) \hat{\pi}(\Bx)

The other required commutator is
\label{eqn:qftLecture10:160}
\begin{aligned}
\antisymmetric{\hatT^{0i}(\Bx)}{\phihat(\By)}
&=
\antisymmetric{\hat{\pi}(\Bx)\partial^i \phihat(\Bx)}{\phihat(\By)} \\
&=
\partial^i \phihat(\Bx)
\antisymmetric{\hat{\pi}(\Bx)
}{\phihat(\By)} \\
&= -i \delta^3(\Bx – \By) \partial^i \phihat(\Bx),
\end{aligned}

The charge commutator with the field can now be computed
\label{eqn:qftLecture10:180}
\begin{aligned}
i \epsilon \antisymmetric{\hatQ^{0k}}{\phihat(\By)}
&=
i
\frac{\epsilon}{2} \int d^3 x
\lr{
x^k
\antisymmetric{\hatT^{00}}{\phihat(\By)}

x^0
\antisymmetric{\hatT^{0k}}{\phihat(\By)}
} \\
&=
\frac{\epsilon}{2} \lr{ y^k \hat{\pi}(\By) – y^0 \partial^k \phihat(\By) } \\
&=
\frac{\epsilon}{2} \lr{ y^k \dot{\phihat}(\By) – y^0 \partial^k \phihat(\By) },
\end{aligned}

so to first order in $$\epsilon$$
\label{eqn:qftLecture10:200}
e^{i \epsilon \hatQ^{0k} } \phihat(\By)
e^{-i \epsilon \hatQ^{0k} }
=
\phihat(\By)
+ \frac{\epsilon}{2} y^k \dot{\phihat}(\By)
+ \frac{\epsilon}{2} y^0 \partial_k \phihat(\By)

For example, with $$k = 1$$
\label{eqn:qftLecture10:700}
\begin{aligned}
e^{i \epsilon \hatQ^{0k} } \phihat(\By)
e^{-i \epsilon \hatQ^{0k} }
&=
\phihat(\By)
+ \frac{\epsilon}{2} \lr{
y^1 \dot{\phihat}(\By)
+
y^0 \PD{y^1}{\phihat}(\By)
} \\
&=
\phihat(y^0 + \frac{\epsilon}{2} y^1,
y^1 + \frac{\epsilon}{2} y^2, y^3).
\end{aligned}

This is a boost. If we compare explicitly to an infinitesimal Lorentz transformation of the coordinates
\label{eqn:qftLecture10:220}
\begin{aligned}
x^0 \rightarrow x^0 + \omega^{01} x_1 &= x^0 – \omega^{01} x^1 \\
x^1 \rightarrow x^1 + \omega^{10} x_0 &= x^1 – \omega^{01} x_0 = x^1 – \omega^{01} x^0
\end{aligned}

we can make the identification
\label{eqn:qftLecture10:240}
\frac{\epsilon}{2} = – \omega^{01}.

We now have the explicit form of the generator of a spacetime translation

\label{eqn:qftLecture10:260}
\boxed{
\hatU(\Lambda) = \exp\lr{-i \omega^{0k} \int d^3 x \lr{ \hatT^{00} x^k – \hatT^{0k} x^0 }}
}

An explicit boost along the x-axis has the form
\label{eqn:qftLecture10:300}
\hatU(\Lambda) \phihat(t, \Bx)
\hatU^\dagger(\Lambda)
=
\phihat\lr{ \frac{t – vx}{\sqrt{1 – v^2}}, \frac{x – vt}{\sqrt{1 – v^2}}, y, z },

and more generally
\label{eqn:qftLecture10:320}
\hatU(\Lambda) \phihat(x) \hatU^\dagger(\Lambda) =
\phihat(\Lambda x)

where $$x$$ is a four vector, $$(\Lambda x)^\mu = {{\Lambda}^\mu}_\nu x^\nu$$, and $${{\Lambda}^\mu}_\nu \approx {{\delta}^\mu}_\nu + {{\omega}^\mu}_\nu$$.

## Transformation of momentum states

In the momentum space representation

\label{eqn:qftLecture10:340}
\begin{aligned}
\phihat(x)
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \lr{
e^{i (\omega_\Bp t – \Bp \cdot \Bx)} \hat{a}_\Bp
+
e^{-i (\omega_\Bp t – \Bp \cdot \Bx)} \hat{a}^\dagger_\Bp
} \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \evalbar{
\lr{
e^{i p^\mu x^\mu } \hat{a}_\Bp
+
e^{-i p^\mu x^\mu } \hat{a}^\dagger_\Bp
}
}{p_0 = \omega_\Bp}
\end{aligned}

\label{eqn:qftLecture10:720}
\begin{aligned}
\hatU(\Lambda) \phihat(x) \hatU^\dagger(\Lambda)
&=
\phihat(\Lambda x) \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \evalbar{
\lr{
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp
+
e^{-i p^\mu {{\Lambda}^\mu}_\nu x^\nu } \hat{a}^\dagger_\Bp
}
}{p_0 = \omega_\Bp}
\end{aligned}

This can be put into an explicitly Lorentz invariant form
\label{eqn:qftLecture10:n}
\begin{aligned}
\phihat(\Lambda x)
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta(p_0^2 – \Bp^2 – m^2) \Theta(p^0) \sqrt{2 \omega_\Bp}
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp + \text{h.c.} \\
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3}
\lr{
\frac{\delta(p_0 – \omega_\Bp)}{2 \omega_\Bp}
+
\frac{\delta(p_0 + \omega_\Bp)}{2 \omega_\Bp}
}
\Theta(p^0) \sqrt{2 \omega_\Bp} \hat{a}_\Bp + \text{h.c.},
\end{aligned}

which recovers \ref{eqn:qftLecture10:720} by making use of the delta function identity $$\delta(f(x)) = \sum_{f(x_\conj) = 0} \frac{\delta(x – x_\conj)}{f'(x_\conj)}$$, since the $$\Theta(p^0)$$ kills the second delta function.

We now have a more explicit Lorentz invariant structure
\label{eqn:qftLecture10:380}
\phihat(\Lambda x)
=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta(p_0^2 – \Bp^2 – m^2) \Theta(p^0) \sqrt{2 \omega_\Bp}
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp + \text{h.c.}

Recall that a boost moves a spacetime point along a parabola, such as that of fig. 1, whereas a rotation moves along a constant “circular” trajectory of a hyper-paraboloid. In general, a Lorentz transformation may move a spacetime point along any path on a hyper-paraboloid such as the one depicted (in two spatial dimensions) in fig. 2. This paraboloid depict the surfaces of constant energy-momentum $$p^0 = \sqrt{ \Bp^2 + m^2 }$$. Because a Lorentz transformation only shift points along that energy-momentum surface, but cannot change the sign of the energy coordinate $$p^0$$, this means that $$\Theta(p^0)$$ is also a Lorentz invariant.

fig. 1. One dimensional spacetime surface for constant (p^0)^2 – p^2 = m^2.

fig. 2. Surface of constant squared four-momentum.

Let’s change variables
\label{eqn:qftLecture10:400}
p^\lambda = {{\Lambda}^\lambda}_\rho {p’}^{\rho}

so that
\label{eqn:qftLecture10:420}
\begin{aligned}
p_\mu
{{\Lambda}^\mu}_\nu x^\nu
&=
{{\Lambda}^\lambda}_\rho {p’}^\rho g_{\lambda\nu} {{\Lambda}^\nu}_\sigma x^{\sigma} \\
&=
{p’}^\rho
\lr{ {{\Lambda}^\lambda}_\rho
g_{\lambda\nu} {{\Lambda}^\nu}_\sigma } x^{\sigma} \\
&=
{p’}^\rho g_{\rho\sigma} x^\sigma
\end{aligned}

which gives
\label{eqn:qftLecture10:440}
\begin{aligned}
\phihat(\Lambda x)
&=
\int \frac{d{p’}^0 d^3 p’}{(2\pi)^3} \delta({p’}_0^2 – {\Bp’}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Lambda \Bp’}} e^{i p’ \cdot x} \hat{a}_{\Lambda \Bp’} + \text{h.c.} \\
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta({p}_0^2 – {\Bp}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Lambda \Bp}} e^{i p \cdot x} \hat{a}_{\Lambda \Bp} + \text{h.c.}
\end{aligned}

Since
\label{eqn:qftLecture10:460}
\phihat(x)
=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta({p}_0^2 – {\Bp}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Bp}} e^{i p \cdot x} \hat{a}_{\Bp} + \text{h.c.}

we can now conclude that the creation and annihilation operators transform as

\label{eqn:qftLecture10:480}
\boxed{
\sqrt{2 \omega_{\Lambda \Bp}} \hat{a}_{\Lambda \Bp}
=
\hatU(\Lambda)
\sqrt{2 \omega_{ \Bp}} \hat{a}_{ \Bp}
\hatU^\dagger(\Lambda)
}

In particular
\label{eqn:qftLecture10:500}
\sqrt{2 \omega_{ \Bp}} \hat{a}^\dagger_{ \Bp} \ket{0} = \ket{\Bp}

and noting that $$\hatU(\Lambda) \ket{0} = \ket{0}$$ (i.e. the ground state is Lorentz invariant), we have
\label{eqn:qftLecture10:520}
\begin{aligned}
\sqrt{2 \omega_{\Lambda \Bp}} \hat{a}^\dagger_{\Lambda \Bp} \ket{0}
&=
\hatU(\Lambda) \sqrt{ 2\omega_\Bp} \hat{a}^\dagger_\Bp \hatU^\dagger(\Lambda) \hatU(\Lambda) \ket{0} \\
&=
\hatU(\Lambda) \sqrt{ 2\omega_\Bp} \hat{a}^\dagger_\Bp \ket{0} \\
&=
\hatU(\Lambda) \ket{\Bp}.
\end{aligned}

## PHY2403H Quantum Field Theory. Lecture 7: Symmetries, translation currents, energy momentum tensor. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Symmetries

Given the complexities of the non-linear systems we want to investigate, examination of symmetries gives us simpler problems that we can solve.

• “internal” symmetries. This means that the symmetries do not act on space time $$(\Bx, t)$$. An example is
\label{eqn:qftLecture7:20}
\phi^i =
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\vdots \\
\psi_N \\
\end{bmatrix}

If we map $$\phi^i \rightarrow O^i_j \phi^j$$ where $$O^\T O = 1$$, then we call this an internal symmetry.
The corresponding Lagrangian density might be something like
\label{eqn:qftLecture7:40}
\LL = \inv{2} \partial_\mu \Bphi \cdot \partial^\mu \Bphi – \frac{m^2}{2} \Bphi \cdot \Bphi – V(\Bphi \cdot \Bphi)

• spacetime symmetries: Translations, rotations, boosts, dilatations. We will consider continuous symmetries, which can be defined as a succession of infinitesimal transformations.
An example from $$O(2)$$ is a rotation
\label{eqn:qftLecture7:60}
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha \\
\end{bmatrix}
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix},

or if $$\alpha \sim 0$$
\label{eqn:qftLecture7:80}
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha \\
-\alpha & 1\\
\end{bmatrix}
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix}
=
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix}
+
\alpha
\begin{bmatrix}
\phi^2 \\
-\phi^1
\end{bmatrix}

In index notation we write
\label{eqn:qftLecture7:100}
\phi^i \rightarrow \phi^i + \alpha e^{ij} \phi^j,

where $$\epsilon^{12} = +1, \epsilon^{21} = -1$$ is the completely antisymmetric tensor. This can be written in more general form as
\label{eqn:qftLecture7:120}
\phi^i \rightarrow \phi^i + \delta \phi^i,

where $$\delta \phi^i$$ is considered to be an infinitesimal transformation.

## Definition: Symmetry

A symmetry means that there is some transformation
\begin{equation*}
\phi^i \rightarrow \phi^i + \delta \phi^i,
\end{equation*}
where $$\delta \phi^i$$ is an infinitesimal transformation, and the equations of motion are invariant under this transformation.

## Theorem: Noether’s theorem (1st).

If the equations of motion re invariant under $$\phi^\mu \rightarrow \phi^\mu + \delta \phi^\mu$$, then there exists a conserved current $$j^\mu$$ such that $$\partial_\mu j^\mu = 0$$.

Noether’s first theorem applies to global symmetries, where the parameters are the same for all $$(\Bx, t)$$. Gauge symmetries are not examples of such global symmetries.

Given a Lagrangian density $$\LL(\phi(x), \phi_{,\mu}(x))$$, where $$\phi_{,\mu} \equiv \partial_\mu \phi$$. The action is
\label{eqn:qftLecture7:160}
S = \int d^d x \LL.

EOMs are invariant if under $$\phi(x) \rightarrow \phi'(x) = \phi(x) + \delta_\epsilon \phi(x)$$, we have
\label{eqn:qftLecture7:180}
\LL(\phi) \rightarrow \LL'(\phi’) = \LL(\phi) + \partial_\mu J_\epsilon^\mu(\phi) + O(\epsilon^2).

Then there exists a conserved current. In QFT we say that the E.O.M’s are “on shell”. Note that \ref{eqn:qftLecture7:180} is a symmetry since we have added a total derivative to the Lagrangian which leaves the equations of motion of unchanged.

In general, the change of action under arbitrary variation of $$\delta \phi$$ of the fields is
\label{eqn:qftLecture7:200}
\begin{aligned}
\delta S
&=
\int d^d x \delta \LL(\phi, \partial_\mu \phi) \\
&=
\int d^d x \lr{
\PD{\phi}{\LL} \delta \phi
+
\PD{(\partial_\mu \phi)}{\LL} \delta \partial_\mu \phi
} \\
&=
\int d^d x \lr{
\partial_\mu \lr{ \PD{(\partial_\mu \phi)}{\LL} } \delta \phi
+
\PD{(\partial_\mu \phi)}{\LL} \partial_\mu \delta \phi
} \\
&=
\int d^d x
\partial_\mu \lr{ \frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta \phi }
\end{aligned}

However from \ref{eqn:qftLecture7:180}
\label{eqn:qftLecture7:220}
\delta_\epsilon \LL = \partial_\mu J_\epsilon^\mu(\phi, \partial_\mu \phi),

so after equating these variations we fine that
\label{eqn:qftLecture7:240}
\delta S = \int d^d x \delta_\epsilon \LL = \int d^d x \partial_\mu J_\epsilon^\mu,

or
\label{eqn:qftLecture7:260}
0 = \int d^d x
\partial_\mu \lr{ \frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta \phi – J_\epsilon^\mu },

or $$\partial_\mu j^\mu = 0$$ provided
\label{eqn:qftLecture7:280}
\boxed{
j^\mu =
\frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta_\epsilon \phi – J_\epsilon^\mu.
}

Integrating the divergence of the current over a space time volume, perhaps that of cylinder (time up, space out) is also zero. That is
\label{eqn:qftLecture7:300}
\begin{aligned}
0
&=
\int d^4 x \, \partial_\mu j^\mu \\
&=
\int d^3 \Bx dt \, \partial_\mu j^\mu \\
&=
\int d^3 \Bx dt \, \partial_t j^0 –
\int d^3 \Bx dt \spacegrad \cdot \Bj \\
&=
\int d^3 \Bx dt \, \partial_t j^0 ,
\end{aligned}

where the spatial divergence is zero assuming there’s no current leaving the volume on the infinite boundary
(no $$\Bj$$ at spatial infinity.)

We write
\label{eqn:qftLecture7:560}
Q = \int d^3x \partial_t j^0,

and call this the on-shell charge associated with the symmetry.

## Spacetime translation.

A spacetime translation has the form
\label{eqn:qftLecture7:320}
x^\mu \rightarrow {x’}^\mu = x^\mu + a^\mu,

\label{eqn:qftLecture7:340}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

(contrast this to a Lorentz transformation that had the form $$x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu$$).

If $$\phi'(x + a) = \phi(x)$$, then
\label{eqn:qftLecture7:360}
\phi'(x) + a^\mu \partial_\mu \phi'(x) =
\phi'(x) + a^\mu \partial_\mu \phi(x) =
\phi(x),

so
\label{eqn:qftLecture7:380}
\phi'(x)
= \phi(x) – a^\mu \partial_\mu \phi'(x)
= \phi(x) + \delta_a \phi(x),

or
\label{eqn:qftLecture7:580}
\delta_a \phi(x) = – a^\mu \partial_\mu \phi(x).

Under $$\phi \rightarrow \phi – a^\mu \partial_\mu \phi$$, we have
\label{eqn:qftLecture7:400}
\LL(\phi) \rightarrow \LL(\phi) – a^\mu \partial_\mu \LL.

Let’s calculate this with our scalar theory Lagrangian
\label{eqn:qftLecture7:420}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi)

The Lagrangian variation is
\label{eqn:qftLecture7:440}
\begin{aligned}
\evalbar{\delta \LL}{\phi \rightarrow \phi + \delta \phi, \delta\phi = – a^\mu \partial_\mu \phi}
&=
(\partial_\mu \phi) \delta (\partial^\mu \phi) – m^2 \phi \delta \phi – \PD{\phi}{V} \delta \phi \\
&=
(\partial_\mu \phi)(-a^\nu \partial_\nu \phi \partial^\mu \phi) + m^2 \phi a^\nu \partial_\nu \phi + \PD{\phi}{V} a^\nu \partial_\nu \phi \\
&=
– a^\nu \partial_\nu \lr{ \inv{2} \partial_\mu \partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi) } \\
&=
– a^\nu \partial_\nu \LL.
\end{aligned}

So the current is
\label{eqn:qftLecture7:600}
\begin{aligned}
j^\mu
&=
(\partial^\mu \phi) (-a^\nu \partial_\nu \phi) + a^\nu \LL \\
&=
-a^\nu \lr{ \partial^\mu \phi \partial_\nu \phi – \LL }
\end{aligned}

We really have a current for each $$\nu$$ direction and can make that explicit writing
\label{eqn:qftLecture7:460}
\begin{aligned}
\delta_\nu \LL
&= -\partial_\nu \LL \\
&= – \partial_\mu \lr{ {\delta^\mu}_\nu \LL } \\
&= \partial_\mu {J^\mu}_\nu
\end{aligned}

we write
\label{eqn:qftLecture7:480}
{j^\mu}_\nu = \PD{x_\mu}{\phi} \lr{ – \PD{x^\nu}{\phi} } + {\delta^\mu}_\nu \LL,

where $$\nu$$ are labels which coordinates are translated:
\label{eqn:qftLecture7:500}
\begin{aligned}
\partial_\nu \phi &= – \partial_\nu \phi \\
\partial_\nu \LL &= – \partial_\nu \LL.
\end{aligned}

We call the conserved quantities elements of the energy-momentum tensor, and write it as
\label{eqn:qftLecture7:520}
\boxed{
{T^\mu}_\nu = -\PD{x_\mu}{\phi} \PD{x^\nu}{\phi} + {\delta^\mu}_\nu \LL.
}

Incidentally, we picked a non-standard sign convention for the tensor, as an explicit expansion of $$T^{00}$$, the energy density component, shows
\label{eqn:qftLecture7:540}
\begin{aligned}
{T^0}_0
&=
-\PD{t}{\phi}
\PD{t}{\phi}
+\inv{2}
\PD{t}{\phi}
\PD{t}{\phi}
– \frac{m^2}{2} \phi^2 – V(\phi) \\
&=
-\inv{2} \PD{t}{\phi} \PD{t}{\phi}
– \frac{m^2}{2} \phi^2 – V(\phi).
\end{aligned}

Had we translated by $$-a^\mu$$ we’d have a positive definite tensor instead.